Question

In Problemsstate the order of the given ordinary differential equation. Determine whether the equation is linear or nonlinear by matching it with.

Short answer

Answer:

The differential equation is of the second order, and the equation is non-linear.

Step-by-step solution

Classification of linearity.

If\(F\)is linear in \(y,y',...,{y^n}\), then the \({n^{th}}\)order ordinary differential equation is said to be linear. The form of the equation is given by,

\({a_n}(x)\frac{{{d^n}y}}{{d{x^n}}} + {a_{n - 1}}(x)\frac{{{d^{n - 1}}y}}{{d{x^{n - 1}}}} + L + {a_1}(x)\frac{{dy}}{{dx}} + {a_0}(x)y = g(x)\)

Determine whether it is linear or nonlinear and state the order

As, by the classification of linearity, the given differential equation should be in the form,\({a_2}(t)\frac{{{d^2}R}}{{d{t^2}}} + {a_1}(t)\frac{{dR}}{{dt}} + {a_0}(t)R = g(t)\).

Square the given equation to get,

\(\begin{aligned}{c}\frac{{{d^2}R}}{{d{t^2}}} + \frac{k}{{{R^2}}} &= 0\\\frac{{{d^2}R}}{{d{t^2}}} + k{R^{ - 2}} &= 0\end{aligned}\)

But the term\(R\) has to be the power of\( - 2\). So, the given is nonlinear.

If the given equation is linear, then the term \(R\)must have to be the power of \(1\). Because of\(\frac{{{d^2}R}}{{d{t^2}}}\), the equation is second order differential equation.